r^2+19r=0

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Solution for r^2+19r=0 equation: 



r^2+19r=0

a = 1; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·1·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*1}=\frac{-38}{2}
=-19
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*1}=\frac{0}{2}
=0
$

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